# Homework Help: The effect of modulus in the graph of a linear equation.

1. Jun 27, 2010

### sharkey1314

My problem is not knowing the effects of modulus in parts of an equation ( in this case linear ) on the parts of the graph. An elaboration will be like y = |bx| + c , whats the effect on the graph compared to y = bx + c and likewise : y = bx + |c| and y = bx + c or even y = b|x| + c compared to the normal y = bx + c. I believe there is a subtle difference. If so what is that because knowing the "relationship" will allow me to solve the problem.

Sorry, i know any genius like Einstein or Newton will have just find out by themselves but not everyone is like them :( My brain can't interpret such complicated -.- or perhaps someone with perseverence...

If possible, explain on this and i will do the question and not just give me the answer. Anyway the question is given below if you are interested.

1. The problem statement, all variables and given/known data
Firstly, i have no idea how to get the image of the graph here. Well, the shape looks like that /\ with the top end linked. y = a - |bx+c|
The follow points are given on the graph : x = -1.5 when y = 4 , x = 0 when y = 3 , x = -1 when y = 3. They both cut the x-axis twice. Find a,b,c and subsequently x-intercepts.

2. Relevant equations
NIL

3. The attempt at a solution
This is "part" of my homework's question but i do not quite understand how modulus affect the graph in various parts of the equation. Therefore, i couldn't attempt the question.

Sharkey

Last edited: Jun 27, 2010
2. Jun 27, 2010

### HallsofIvy

Look at what is inside the "| |". Where is that equal to 0? The change in the graph will occur there. For example, if y= a- |bx+ c| then we look at bx+ c= 0 or x= -c/b. For x< -c/b, bx< c so bx+c< 0. |bx+c|= -(bx+ c) and so y= a- |bx+c|= y= a+ bx+ c, a straight line with slope b.

if x> -c/b, then bx+c> 0 and |bx+c|= bx+ c so y= a- bx- c, a straight line with slope -b.

In this problem, since we don't know what b and c are, we have treat it in cases.

Since a straight line is "one-to-one" and here we have y(0)=3, y(-1)= 3, the "break" must occur between -1 and 0. That is -1< -c/b< 0. Both -1.5 and -1 are less than -c/b so we have 4= a+ b(-1.5)+ c, 3=a+ b(-1)+ c. 0 is larger than -c/b so 3= a- b(0)- c.

That gives three equations to solve for a, b, and c.

3. Jun 27, 2010

### sharkey1314

I'm abit confused. Isn't modulus is about making what is inside | | positive ? then y= a- |bx+c| should result in y = a - bx -c ? Why there is y = a + bx + c? Perhaps my understanding of modulus isn't that strong lol.. but my teacher told us modulus means regarding it is positive or negative, it will become positive eventually.

4. Jun 27, 2010

### ehild

|x|= x if x≥0; |x|= -x if x<0.

|x| is never negative.

So |2|=2 and |-2|=2.

|bx+c|=bx+c if x≥-c/b, and |bx+c|=-(bx+c) if x<-c/b.

ehild

5. Jun 27, 2010

### HallsofIvy

Yes, it is about making what is inside | | positive. It doesn't mean that what is inside is already positive which is what you are trying to say. For example, in y= 2- |x+ 3|, if x= -4, then x+ 3= -4+ 3= -1< 0 so |x+3|= -(-1)= 1. y= 2- |x+3| becomes y= 2-1= 1 for x= -4. That is different from simply y= 2- x- 3= -x- 1 which is 0 for x= -4.

As ehild said, if $x\ge 0$, then |x|= x but if x< 0, then |x|= -x.

Here, if x> -3, x+3> 0 so |x+3|= x+3 and y= 2- |x+3| becomes y= 2-x-3= -x-1. But if x< -3, |x+3|= -(x+3)= -x-3 and y= 2-|x+3| becomes y= 2-(-x-3)= x+ 5.

The graph of y= -x-1 is a straight line with slope -1 so is a descending line.
Of course, that is the graph for this function only for x< -3.

The graph of y= x+ 5 is a straight line with slope 1 so is an increasing line.
That is the graph of this function for x> -3.

If x= -3, x+ 3= 0 so |x+ 3|= 0 and y= 2- |x+3|= 2.

Notice that if x= -3, y= -x-1 becomes y=-(-3)-1 = 3-1= 2 and if y= x+5= -3+ 5= 2 so the two line meet at (-3, 2). That's what gives the "upside down v" shape to the entire graph.

6. Jun 27, 2010

### sharkey1314

I get it. So essentially modulus means there isthree possibilities : negative, positive and zero. For negative, the modulus will be multiplied (-1) to make it positive whereas positive will be left as it is. Zero means the meeting point ? <== is it alwaysthe same case for other kind of equations like quadratic, cubic.

In summary, to approach such questions, have to form three possible equations and using these equations to solve ?

7. Jun 27, 2010

### ehild

Yes, but it is only two cases, as |0|=0.

For example the function y=|x-1|is equivalent to

y=-x+1 if x<1 and y=x-1 if x≥1.

Or y=|x^2| is equivalent to y=x^2, as x^2 can not be negative, but

y=|x^3| is equivalent to y=-x^3 if x<0 and y=x^3 if x ≥0.

If you have something like y=|bx| , it is y=-bx if bx<0, that is when b and x are of opposite signs. Otherwise, when both b and x are positive or negative, bx>0 and |bx|=bx.

ehild